Is the "op" lattice ##\mathscr{L_H}^\perp## also atomistic....?

[forkosh]

TL;DR

Is the "op" lattice of subspaces of a Hilbert space also atomistic with the covering property?

Let $\mathscr{L_H}$ be the usual lattice of subspaces of Hilbert space $\mathscr{H}$, where for $p,q\in\mathscr{H}$ we write $p\leq q$ iff $p$ is a subspace of $q$. Then, as discussed by, e.g., Beltrametti&Cassinelli https://books.google.com/books?id=yWoq_MRKAgcC&pg=PA98, this lattice is atomistic with the covering property. That is, enormously briefly, $0$ is the weakest element, i.e., $\forall q\in\mathscr{H}: 0\leq q$, and pure states $p$ (one-dimensional subspaces) are "atoms" with the covering property, i.e., $\forall q\in\mathscr{H}: 0\leq q\leq p\Longrightarrow 0=q\mbox{ .or. }q=p$.

Now consider the "op" lattice $\mathscr{L_H}^\perp$ where $q\leq p$ in $\mathscr{L_H}$ means $p\leq q$ in $\mathscr{L_H}^\perp$. Then $0^\perp=\mathscr{H}\in \mathscr{L_H}^\perp$ is the weakest element. And can we say that (a) the orthocomplements of pure states are the atoms of $\mathscr{L_H}^\perp$, and that (b) they also possesses the corresponding covering property with respect to this lattice? And can you prove it, or even better (and presumably easier) cite a proof? Note that by "prove", you can assume it's true for $\mathscr{L_H}$, and only need to prove it's then also true for $\mathscr{L_H}^\perp$.

P.S. Another discussion of atomistic lattices with the covering property is in Section 4.2.5 (pages 4-10 and 4-11) of https://arxiv.org/abs/1211.5627

 

[azntoon]

.Yes, it is true that orthocomplements of pure states are the atoms of $\mathscr{L_H}^\perp$, and they also possess the covering property with respect to this lattice. To prove this, we will first show that the orthocomplements of pure states are the atoms of $\mathscr{L_H}^\perp$, and then show that they have the covering property.First, we will show that the orthocomplements of pure states are the atoms of $\mathscr{L_H}^\perp$. Suppose that $p \in \mathscr{L_H}$ is a pure state, and let $p^\perp \in \mathscr{L_H}^\perp$ be its orthocomplement. We want to show that $p^\perp$ is an atom in $\mathscr{L_H}^\perp$. That is, we need to show that for any two subspaces $q_1, q_2 \in \mathscr{L_H}^\perp$ such that $q_1 \leq p^\perp \leq q_2$, either $q_1 =0$ or $q_2 = p^\perp$.Let $q_1, q_2 \in \mathscr{L_H}^\perp$ be two subspaces such that $q_1 \leq p^\perp \leq q_2$. By definition, this means that $p \leq q_1^\perp \leq p^\perp \leq q_2^\perp$. Now, since $p$ is a pure state, it follows from the atomicity of $\mathscr{L_H}$ that either $q_1^\perp =0$ or $q_2^\perp = p$. This implies that either $q_1 =0$ or $q_2 = p^\perp$, as desired. Thus, we have shown that the orthocomplements of pure states are the atoms of ##\mathscr{L_H}^